Question

Find $\underset{x\to 0}{lim}\frac{1-\cos 2x+{\tan }^{2}x}{x\sin x}$

Answer 1

$\underset{x\to 0}{lt}\frac{1-\cos 2x+{\tan }^{2}x}{x\sin x}$

$=\underset{x\to 0}{lt}\frac{2{\sin }^{2}x+{\tan }^{2}x}{x\sin x}$

$=\underset{x\to 0}{lt}\frac{2\sin x}{x}+\frac{{\sin }^{2}x}{x\sin x{\cos }^{2}x}$

$=\underset{x\to 0}{lt}\frac{2\sin x}{x}+\underset{x\to 0}{lt}\cdot \frac{\sin x}{x}\cdot \frac{1}{{\cos }^{2}x}$

$=2+1=3$.